Question:

Step 1: Rearrange and separate variables We start with the differential equation: $$e^{-x}(y+1)\,dy + (\cos^2 x - \sin 2x)y\,dx = 0$$ Rearranging to isolate the differentials: $$e^{-x}(y+1)\,dy = -(\cos^2 x - \sin 2x)y\,dx$$ Dividing both sides by $ye^{-x}$: $$\frac{y+1}{y}\,dy = -e^x(\cos^2 x - \sin 2x)\,dx$$ This can be rewritten as: $$\left(1 + \frac{1}{y}\right)dy = -e^x(\cos^2 x - \sin 2x)\,dx$$ Step 2: Integrate the left-hand side We integrate the left side: $$\int\left(1 + \frac{1}{y}\right)dy = y + \ln|y| + C_1$$ Step 3: Integrate the right-hand side For the right side, we need to evaluate: $$-\int e^x(\cos^2 x - \sin 2x)\,dx$$ We recognize that this is related to the derivative of $e^x\cos^2 x$. Using the product rule: $$\frac{d}{dx}(e^x\cos^2 x) = e^x\cos^2 x + e^x \cdot \frac{d}{dx}(\cos^2 x)$$ Since $\frac{d}{dx}(\cos^2 x) = 2\cos x(-\sin x) = -\sin 2x$, we have: $$\frac{d}{dx}(e^x\cos^2 x) = e^x\cos^2 x - e^x\sin 2x = e^x(\cos^2 x - \sin 2x)$$ Therefore: $$-\int e^x(\cos^2 x - \sin 2x)\,dx = -e^x\cos^2 x + C_2$$ Step 4: Combine both sides and simplify Equating the integrals from both sides: $$y + \ln|y| = -e^x\cos^2 x + C$$ Rearranging: $$y + \ln y + e^x\cos^2 x = C$$ Step 5: Apply the initial condition We use the condition $y = 1$ when $x = 0$: $$1 + \ln(1) + e^0\cos^2(0) = C$$ $$1 + 0 + 1 \cdot 1 = C$$ $$C = 2$$ Step 6: State the final solution The solution to the differential equation is: $$y + \ln y + e^x\cos^2 x = 2$$ This matches **Option 4**. <div class="key-concept"><strong>Key Concept:</strong> Separable differential equations, integration by recognition of derivative</div> <div class="trap-box"><strong>Trap:</strong> Recognizing that $\frac{d}{dx}(e^x\cos^2 x) = e^x(\cos^2 x - \sin 2x)$ to evaluate the RHS integral.</div>